From the postion you posted, I would note that since 7 is already placed in box 1 you can eliminate it from the possibilities for r2c2. This then leads to a triple {3,6,8} in r1c2, r2c2, r3c2. Therefore there is a pair {1,2} in r1c1, r3c1. Also there is then a pair {1,2} in r3c1, r3c6, which reduces the possibilities for r3c9. Then looking at box 3 you will have another triple (and check the 8 which is already placed but still appears as a candidate).

Thank you, ravel, for responding, but I don't see the hidden pair -- yet. Could you point out the cells that constitute the interaction?

The way, Angel did it - spotting the triple, which gives the pair, is a bit easier, but i did not see the triple because of the 7.

The 1 in box 1 can only be in column 1 (therefore must be there), so it cannot be in the rest of the column (box/line interaction) and it can be eliminated from r5c1 and r6c1. This leaves the hidden pair 12 in r13c1.
Sorry, there is no box/box elimination for the same.

PS: I dont really understand the hint r4c6=9. i would need the following chain to get that:
r4c6=6 => r5c6=9 => r6c3=9 => r7c9=9 => r4c6=9
which is very complicated compared to a naked triple.